Angioplasty is a medical procedure in which an obstructed blood vessel is widened. In some cases, a wire mesh tube, called a stent, is placed in the vessel to help it remain open. A study was conducted to compare the effectiveness of a bare metal stent with one that has been coated with a drug designed to prevent reblocking of the vessel. A total of 5326 patients received bare metal stents, and of these, 848 needed treatment for reblocking within a year. A total of 1121 received drug-coated stents, and 141 of them required treatment within a year. Can you conclude that the proportion of patients who needed retreatment differs between those who received bare metal stents and those who received drug-coated stents? Let pv1 denote the proportion of patients with bare metal stents who needed retreatment and pv2 denote the proportion of patients with drug-coated stents who needed retreatment. Use the a=0.10 level of significance and the P-value method

Question

State the appropriate null and alternate hypothesis.

Answer 1

To analyze whether there is a difference in the proportion of patients who needed retreatment between those who received bare metal stents and those who received drug-coated stents, we can state the null and alternative hypotheses as follows:

Null Hypothesis (H0):
There is no difference in the proportion of patients needing retreatment between the two groups. In symbolic terms, this can be expressed as: \( H_0: p_1 = p_2 \)
where \( p_1 \) is the proportion of patients needing retreatment with bare metal stents (pv1) and \( p_2 \) is the proportion of patients needing retreatment with drug-coated stents (pv2).

Alternative Hypothesis (H1):
There is a difference in the proportion of patients needing retreatment between the two groups. This can be expressed as: \( H_1: p_1 \neq p_2 \)

In conclusion:

• \( H_0: p_1 = p_2 \)
• \( H_1: p_1 \neq p_2 \)

This represents a two-tailed test since we are checking for any difference, not specifying a direction (greater or lesser). The significance level for the test is \( \alpha = 0.10 \).

Answer 2

Find the P-value. Round the answer to at least four decimal places.

Answer 3

To calculate the P-value for the hypothesis test comparing the two proportions, we can use the following steps:

Step 1: Calculate the sample proportions

1. For bare metal stents: \[ p_1 = \frac{848}{5326} \approx 0.159 \]

2. For drug-coated stents: \[ p_2 = \frac{141}{1121} \approx 0.126 \]

Step 2: Calculate the pooled proportion

The pooled proportion (\( p \)) can be calculated using the formula: \[ p = \frac{x_1 + x_2}{n_1 + n_2} \] where \( x_1 \) and \( x_2 \) are the number of successes (retreatments), and \( n_1 \) and \( n_2 \) are the sample sizes for each group.

• \( x_1 = 848, , n_1 = 5326 \)
• \( x_2 = 141, , n_2 = 1121 \)

Calculating the pooled proportion: \[ p = \frac{848 + 141}{5326 + 1121} = \frac{989}{6447} \approx 0.1534 \]

Step 3: Calculate the standard error (SE)

The standard error for the difference in proportions is calculated as: \[ SE = \sqrt{p(1 - p) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \]

Substituting the values:

• \( p \approx 0.1534 \)
• \( n_1 = 5326 \)
• \( n_2 = 1121 \)

Calculating \( SE \): \[ SE = \sqrt{0.1534 \times (1 - 0.1534) \left( \frac{1}{5326} + \frac{1}{1121} \right)} \] \[ = \sqrt{0.1534 \times 0.8466 \left( \frac{1}{5326} + \frac{1}{1121} \right)} \] Calculating the fractions: \[ = \sqrt{0.1534 \times 0.8466 \left( 0.0001875 + 0.0008916 \right)} \] \[ = \sqrt{0.1534 \times 0.8466 \times 0.0010791} \approx \sqrt{0.00013331} \approx 0.0115 \]

Step 4: Calculate the Z-score

The Z-score for the difference in proportions is calculated as: \[ Z = \frac{p_1 - p_2}{SE} \]

Substituting the values: \[ = \frac{0.159 - 0.126}{0.0115} \approx \frac{0.033}{0.0115} \approx 2.87 \]

Step 5: Calculate the P-value

Now we can find the P-value using the Z-score. We will look up the Z-score in a standard normal distribution table or use a calculator or software.

For \( Z \approx 2.87 \), we can find the P-value for a two-tailed test:

• The area to the right of \( Z = 2.87 \) is approximately \( 0.0021 \).
• Therefore, for a two-tailed test, we multiply by 2: \[ P\text{-value} \approx 2 \times 0.0021 = 0.0042 \]

Final Answer

The P-value is approximately:

\[ \text{P-value} \approx 0.0042 \]

Rounding

Rounding to four decimal places gives us:

\[ \boxed{0.0042} \]

Answer 4

Do we reject Hv0?

Answer 5

To determine whether to reject the null hypothesis (\(H_0\)) based on the calculated P-value, we can compare the P-value to the significance level \(\alpha\).

Given Data:

• P-value: \(0.0042\)
• Significance level (\(\alpha\)): \(0.10\)

Decision Rule:

• If \( \text{P-value} < \alpha \), we reject the null hypothesis.
• If \( \text{P-value} \geq \alpha \), we do not reject the null hypothesis.

Comparison:

Since \(0.0042 < 0.10\), we reject the null hypothesis (\(H_0\)).

Conclusion:

This suggests that there is a statistically significant difference in the proportion of patients needing retreatment between those who received bare metal stents and those who received drug-coated stents at the 0.10 level of significance.

Answer 6

There [is or is not?] enough evidence to conclude that the proportion of patients who needed retreatment differs between those who received bare metal stents and those who received drug-coated stents.

Answer 7

There is enough evidence to conclude that the proportion of patients who needed retreatment differs between those who received bare metal stents and those who received drug-coated stents.

This conclusion is based on the fact that we rejected the null hypothesis (\(H_0\)) at the 0.10 significance level, indicating a statistically significant difference in the proportions.